Question

If $\frac{x^2+1}{x}=3\frac{1}{3}$ and $x>1$, find the value of $x-\frac{1}{x}$.

• A. $2\frac{2}{3}$
• B. $1\frac{2}{9}$
• C. $2\frac{5}{7}$
• D. $1\frac{1}{3}$

Answer 1

The correct option is A $2\frac{2}{3}$

$\frac{x^2+1}{x}=3\frac{1}{3}$
$=>x+\frac{1}{x}=\frac{10}{3}$
Squaring both sides,
$(x+\frac{1}{x}{)}^2=(\frac{10}{3}{)}^2$
$=>x^2+(\frac{1}{x}{)}^2+2(x\left)\right(\frac{1}{x})=\frac{100}{9}$
$=>x^2+(\frac{1}{x}{)}^2=\frac{100}{9}-2$
$=>x^2+(\frac{1}{x}{)}^2=\frac{82}{9}$
Now,
$(x-\frac{1}{x}{)}^2=x^2+(\frac{1}{x}{)}^2-2\left(x\right)\left(\frac{1}{x}\right)$
$=(\frac{82}{9}-2)$
$=\left(\frac{64}{9}\right)$
$(x-\frac{1}{x})=\sqrt{(\frac{64}{9}})$
$=\left(\frac{8}{3}\right)$
$=2\frac{2}{3}$